Question 279618
.
Investor Company loaned out a total of $36,000, part at 6% interest and part at 9% interest. 
They reported that the annual earnings from both investments were the same amount that would have been earned 
by the total loan if it had been invested at 8%. Find the amount loaned at each rate.
~~~~~~~~~~~~~~~~~~~~~~~~~~



        The solution in the post by @mananth is incorrect,

        since the governing equation was setup incorrectly in his post.


        I came to bring a correct and accurate solution.



<pre>
Let x be the amount invested at 9% interest.
Then (36000-x) dollars invested at 6%.


Write the total interest equation

    0.09x + 0.06*(36000-x) = 0.08*36000.


Simplify and find x

    0.09x + 0.06*36000 - 0.06x = 0.08*36000,

    0.09x - 0.06x = 0.08*36000 - 0.06*36000,

    0.03x = 0.02*36000

        x = {{{(0.02*36000)/0.03}}} = 24000.


<U>ANSWER</U>.  $24000 was invested at 9% and the rest, 36000-24000 = 12000 dollars, was invested at 6%.


<U>CHECK</U>.   The total interest is  0.09*24000 + 0.06*12000 = 2880 dollars.

         Calculated by another way, it is  0.08*36000 = 2880 follars, the same amount.

         The solution is confirmed to be correct.
</pre>

Solved correctly.